Elementary Analysis Math 140B—Winter 2007
                            Homework answers—Assignment 11; February 19, 2007
                                                              2
                    Define f as follows: f(x) = exp(−1/x ) for x 6= 0, and f(0) = 0. Show that
                (a) f is continuous for all x.
                       Solution: Since f is differentiable everywhere (by (b), which doesn’t use (a)
                       in its proof), f is continuous.
                (b) The n-th derivative of f is continuous for all x and vanishes at x = 0, for all
                       n≥1.
                       Solution: The only statement that needs proof is f(k)(0) = 0,
                       NOTE:This solution is taken from R. P. Boas, A Primer of Real Functions, p.
                       154. Another solution is given as the solution to Exercise 31.5 in the book by
                       Ross. (I prefer the solution in Boas.)
                                                                                  (k)              −1/x2
                         1. It is easy to see by induction that for x 6= 0, f        (x) = R(x)e         where
                             Ris a rational function, R(x) = P(x)/Q(x) for polynomials P and Q.
                                                                                    n −1/x2
                         2. For any integer n, positive, negative, or zero, x e              →0as x → 0.
                                                   −1/x2                     n             −1/x2
                             Indeed, for n = 0, e        →0; for n > 0, x → 0 and e               ≤1; and for
                                                                 2           2
                                                          n −1/x       −n   t
                             n<0,let t = 1/x to get x e            =t /e →0ast→∞byL’Hopital’s
                             rule.
                         3. LEMMA: If lim          f′(x) exists as a finite real number, then f′(y) exists
                                               x→y
                             and equals this limit.
                             PROOF: By the mean value theorem,
                                           f(x)−f(y) =f′(t) for some t between x and y.                     (1)
                                               x−y
                             As x approaches y, so does t, so the right side of (1) approaches a limit.
                             By the left side of (1), this limit is f′(y).
                         4. By 1. and 2. , f(k)(x) → 0 as x → 0, hence by 3. , f(k)(0) = 0. To prove
                             that f(k)(x) → 0, write
                                               P(x)        2    a xm+a        xn−1 +···+a              2
                                     (k)               −1/x       n        n−1                0    −1/x
                                   f    (x) = Q(x) e         =b xm+b          xm−1 +···+b        e      .
                                                                 m        m−1                 0
                             If b0 6= 0, then
                                                             m          n−1               −1/x2
                                                lim     (a x +a        x     +···+a )e              0
                                 limf(k)(x) =       x→0 n          n−1                 0         =     =0.
                                 x→0               lim     (b xm+b        xm−1 +···+b )             b
                                                       x→0 m          m−1                  0         0
                             If b0 = 0, write Q(x) = xk(bmxm−k +···+bk), where bk 6= 0. Then
                                                        −k     m           n−1              −1/x2
                                              lim     x (a x +a          x     +···+a )e              0
                               limf(k)(x) =       x→0        n        n−1                0         =     =0.
                               x→0               lim    (b xm−k +b        xm−1−k +···+b )             b
                                                    x→0 m             m−1                    k         k
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